Flynn had three bags of black beans, E, F and G and the mass of each bag of black beans was in the ratio 2 : 3 : 4. Flynn decided to transfer 40% of black beans from Bag E into Bag F and 90% of black beans from Bag G into Bag F. Given that the mass of Bag F was 59.2 kg in the end, how many kilograms of black beans were transferred into Bag F?

	E	F	G
Before	2 u	3 u	4 u
Change 1	- 0.8 u	+ 0.8 u
Change 2		+ 3.6 u	- 3.6 u
After	1.2 u	7.4 u	0.4 u

Mass of black beans transferred from Bag E into Bag F
= 40% x 2 u
= ⁴⁰₁₀₀ x 2 u
= 1.2 u

Mass of black beans transferred from Bag G into Bag F
= 90% x 4 u
= ⁹⁰₁₀₀ x 4 u
= 3.6 u

Mass of black beans in Bag F in the end
= 3 u + 0.8 u + 3.6 u
= 7.4 u

7.4 u = 59.2 

1 u = 59.2 ÷ 7.4 = 8

Mass of black beans transferred into Bag F
= 0.8 u + 3.6 u
= 4.4 u
= 4.4 x 8
= 35.2 kg

Answer(s): 35.2 kg